Computer Engineering 4TL4: Digital Signal Processing (Fall 2003) Solutions to Final Exam

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1 Computer Engineering TL: Digital Signal Proessing (Fall 3) Solutions to Final Exam The step response ynof a ausal, stable LTI system is: n [ ] = [ yn ] un, [ ] where un [ ] is the unit step funtion a Find the -domain transfer funtion H( ) of this system b Find the impulse response hn [ ] of this system Find the linear onstant oeffiient differene (LCCD) equation that desribes this system a Taking the -transform of yn [ ] gives: Y =, > The -transform X ( ) of the input unit step funtion is: X =, > The transfer funtion H( ) is then: Y = =, X > b The impulse response of this system is found by omputing the inverse -transform of H( ) : = = n n [ ] [ ] [ ] hn= un un The LCCD equation that desribes this system is: Y ( ) = = X ( ) Y ( ) Y ( ) X ( ) X ( ) yn [ ] yn [ ] xn [ ] xn [ ] = = >

2 Consider the ausal stable LTI system with the transfer funtion: = ( )( + + ) Draw the blok diagram for this system implemented in parallel form with st -order subsystems To obtain the parallel form, we ompute the partial fration expansion of H( ) : = = ( )( + + ) + + The blok diagram is then: 3 A ausal LTI system has the transfer funtion: = ( + )( 9 ) + 8 a Find transfer funtions for a minimum-phase system and an all-pass system Hap ( ) suh that: = ap b Sketh the pole-ero plots of, and, learly show their ROCs by shading them on the pole-ero plots H( ) H ( ) ap

3 a The system is nonminimum-phase beause of the pair of eros at =±3 To form the minimum-phase system, these eros must be refleted to their (onjugate) reiproal loations at H ( ) = =± 3, giving: ( + 9 ) + 8, > 9 H( ) Hap( ) H( ) ap =± In order for to equal, the all-pass system must have the pair of nonminimum-phase eros at 3 from and a pair of poles at the reiproal loations =± 3, giving: H ( ) 9 =, > ap 9 H( ) H ( ) b The pole-ero plots of, and, with their ROCs, are given below ap 3 Imaginary Part Imaginary Part Imaginary Part H() ROC -plane 3 3 Real Part H () ROC -plane 3 3 Real Part H ap () ROC -plane 3 3 Real Part

4 The spetrogram of a speeh signal is shown below a Is this a narrowband spetrogram or a wideband spetrogram? Explain your reasoning b This signal is a single word, either shop or posh From the spetrogram, whih of these two words does it appear to be? Explain your reasoning 8 Frequeny (kh) 3 Time (s) a This is a wideband spetrogram beause the temporal resolution is good and the frequeny resolution is poor, suh that vertial striations are seen in the voied segment of speeh and the harmonis are unresolved (ie, no horiontal striations are observed) b The word is shop, whih is evident in the spetrogram beause of the long noise-like spetrum at the beginning, produed by the unvoied friative sh Two alternative methods of digital IIR filter design make use of the: i impulse invariane transformation and ii bilinear transformation Explain: a whih of these two methods is appliable to a wider range of filter frequeny response types (ie, lowpass, highpass, bandpass, bandstop, multiband) and why, and b whih of these two methods is simpler and why a The bilinear transform is appliable to a wider range of filter frequeny responses beause it maps the entire analog frequeny axis into the digital frequeny range ( π, π), suh that frequeny aliasing is avoided The impulse invariane method does not warp the frequeny axis, so aliasing is produed exept in ases of stritly bandlimited frequeny responses b The impulse invariane method is somewhat simpler beause it just requires sampling and saling the impulse response of the desired analog filter However, there is no simple mapping of the s-domain transfer funtion to the -domain In onstrast, the bilinear transform requires prewarping of the desired digital filter properties before design of the analog filter, appliation of the bilinear transformation and simplifiation of the -domain transfer funtion, but there is a diret mapping of the s-domain transfer funtion to the -domain

5 Four values of a stationary random proess x[n] have been observed, giving the disrete-time sequene v[n] = {38,, 7, 8} a Use the autoorrelation-based method of spetral estimation to estimate the PSD of x[n], using a triangular orrelation window w [m] = {,, } b Does x[n] appear to be lowpass, bandpass or highpass? a First we ompute the estimate of autoorrelation funtion of the random proess Note that we will be applying a orrelation window with nonero values only at lags of m =, and, so we only need to evaluate the autoorrelation estimator at those lags: [ ] [ ] [ ] m = v m v m [ ] [ ] [] [ ] = = 9 = = 88 = = 9 ˆ φ [ m] = [ m] = [ m] Q Applying the triangular orrelation window gives: ˆ φ 9δ 88δ 9δ δ δ = [ m] w [ m] = { [ m+ ] + [ m] + [ m ]} { [ m+ ] + [ m] + δ[ m ]} { δ[ m ] δ[ m] δ[ m ]} Taking the Fourier transform of the windowed autoorrelation estimate gives the PSD estimator: ( ω) = ˆ φ [ ] [ ] S m w m e m= m= jωm { 3δ[ m ] 7δ [ m] 3δ [ m ] } jω = 3e e = 7+ 7os ω = jω e jωm b From the final expression for S ( ω ) we would estimate that x[ n ] is a lowpass signal

6 7 Let h [ ] lp n denote the impulse response of an ideal lowpass filter with unity passband gain and utoff frequeny filter [ ] ω = π The figure below shows an ideal LTI frequeny-seletive j h [ ] H e ω lp n hn that inorporates as a subsystem Sketh, the magnitude frequeny response of hn, [ ] indiating expliitly the band-edge frequenies in terms of and speify whether the system is a lowpass, highpass, bandpass, bandstop or multiband filter ω () n () n x[n] h lp [n] y[n] h[n] x n bandlimited to ( π, π), shown by the blue line in panel A of the figure below Multipliation of x[ n ] by the sequene ( ) n in the time domain orresponds to a frequeny shift of π radians in the frequeny domain, from Property 3 of Table, where ω = π in this ase This frequeny shift turns the lowpass signal x[ n] into a highpass signal, shown in panel B After lowpass filtering (shown by the red dashed line in panel B), the signal (shown in panel C) is again frequeny shifted by π radians to give the blue line in panel D The net filtering effet of these operations is a highpass filter with utoff frequeny ω = π ω = π π = 3π, as illustrated by the magenta line in panel D Consider a lowpass disrete-time signal [ ] A Magnitude D Magnitude or Gain 3π π π π π 3π ω (radians) ω ' = 3π/ 3π π π π π 3π ω (radians) Frequeny shift Frequeny shift B Gain Magnitude or C Magnitude ω = π/ 3π π π π π 3π ω (radians) Filter 3π π π π π 3π ω (radians)

7 8 Consider the finite sequene xn [ ] = {, 7, 9,,, 8} The DFT of this sequene X [ k ] has been alulated for k and lossy ompression has been applied to the sequene of DFT oeffiients X [ k ] by ignoring (ie, setting to ero) the oeffiients for k =, 3 and, to obtain the sequene of ompressed DFT oeffiients: [ ] = { 7, 39,,,, + 39} X k j j a Compute the reonstruted sequene [ n] by taking the inverse DFT of X [ k] b Calulate the root mean squared (RMS) error between [ n] and [ ] RMS error as a perentage of the RMS of x[ n ] a The inverse DFT of X [ k] is: jπ kn = [ ] [ ] xn k = [ ] = [ ] [] = [ ] [ ] = [ ] k = k = k = X ke X k e = { 7 + j j39} = 77 jπ k X k e jπ jπ { 7 ( j39) e ( j ) e } jπ jπ = { 7 + ( j39) e + ( + j39) e } = 9 jπ k X k e = + jπ k3 = X k e [ 3] [ ] k = [ ] = [ ] k = [ ] = [ ] = 7 x n Express this jπ3 jπ = { 7 ( 7 + j837) e + ( + j39) e } = jπ k X k e jπ jπ = { 7 ( 7 + j837) e + ( + j39) e } = 983 jπ k X k e k = jπ jπ = { 7 + ( j39) e + ( + j39) e } =, giving the sequene xn= [ ] { 77, 9, 7,, 983, }

8 b The RMS error is: {( [ ] [ ]) } RMSE = E x n x n The RMS of x[ n ] is: RMS= ( 77) + ( 7 9) + ( 9 7) = + ( + ) + ( + 983) + ( 8 + ) = 38 E {( x[ n] ) } { } = = 7, giving a perentage RMS error of: RMSE 38 = 8% RMS 7